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a) Ta có: \(3\sqrt{2}+4\sqrt{8}-\sqrt{18}\)
\(=\sqrt{2}\left(3+4\cdot2-3\right)\)
\(=8\sqrt{2}\)
b) Ta có: \(\sqrt{3}-\frac{1}{3}\sqrt{27}+2\sqrt{507}\)
\(=\sqrt{3}\left(1-\frac{1}{3}\cdot\sqrt{9}+2\cdot\sqrt{169}\right)\)
\(=\sqrt{3}\left(1-1+26\right)\)
\(=26\sqrt{3}\)
c) Ta có: \(\sqrt{25a}+\sqrt{49a}-\sqrt{64a}\)
\(=\sqrt{25}\cdot\sqrt{a}+\sqrt{49}\cdot\sqrt{a}-\sqrt{64}\cdot\sqrt{a}\)
\(=\sqrt{a}\left(5+7-8\right)\)
\(=4\sqrt{a}\)
d) Ta có: \(-\sqrt{36b}-\frac{1}{3}\sqrt{54b}+\frac{1}{5}\sqrt{150b}\)
\(=-\sqrt{6b}\cdot\sqrt{6}-\frac{1}{3}\cdot\sqrt{6b}\cdot\sqrt{9}+\frac{1}{5}\cdot\sqrt{6b}\cdot\sqrt{25}\)
\(=-\sqrt{6b}\left(\sqrt{6}+1-1\right)\)
\(=-\sqrt{6b}\cdot\sqrt{6}=-6\sqrt{b}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(3+1\sqrt{6}-\sqrt{33}\right)\left(\sqrt{22}+\sqrt{6}+4\right)\)
\(=\sqrt{3}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right).\sqrt{2}\left(\sqrt{11}+\sqrt{3}+2\sqrt{2}\right)\)
\(=\sqrt{6}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right)\left(\sqrt{3}+2\sqrt{2}+\sqrt{11}\right)\)
\(=\sqrt{6}\left[\left(\sqrt{3}+2\sqrt{2}\right)^2-11\right]=\sqrt{6}\left(11+4\sqrt{6}-11\right)=\sqrt{6}.4\sqrt{6}=6.4=24\)
b) \(\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)=\left(\frac{5+2\sqrt{6}+10-4\sqrt{6}}{5^2-\left(2\sqrt{6}\right)^2}\right)\left(15+2\sqrt{6}\right)\)
\(=\left(15-2\sqrt{6}\right)\left(15+2\sqrt{6}\right)=15^2-24=201\)
C) \(\left(\frac{4}{3}.\sqrt{3}+\sqrt{2}+\sqrt{3\frac{1}{3}}\right)\left(\sqrt{1,2}+\sqrt{2}-4\sqrt{\frac{1}{5}}\right)\)
\(=\left(\frac{4}{\sqrt{3}}+\frac{\sqrt{6}}{\sqrt{3}}+\frac{\sqrt{10}}{\sqrt{3}}\right)\left(\frac{\sqrt{6}}{\sqrt{5}}+\frac{\sqrt{10}}{\sqrt{5}}-\frac{4}{\sqrt{5}}\right)\)
\(=\frac{1}{\sqrt{15}}\left(\sqrt{6}+\sqrt{10}+4\right)\left(\sqrt{6}+\sqrt{10}-4\right)=\frac{1}{\sqrt{15}}\left[\left(\sqrt{6}+\sqrt{10}\right)^2-16\right]\)
\(=\frac{1}{\sqrt{15}}\left(16+4\sqrt{15}-16\right)=\frac{4\sqrt{15}}{\sqrt{15}}=4\)
d) \(\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1990+2\sqrt{1989}}=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1989+2\sqrt{1989}+1}\)
\(=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{\left(\sqrt{1989}+1\right)^2}=\left(\sqrt{1989}-1\right)\left(\sqrt{1989}+1\right)=1989-1=1988\)
e) \(\frac{a-\sqrt{ab}+b}{a\sqrt{a}+b\sqrt{b}}-\frac{1}{a-b}=\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}-1}{a-b}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(\sqrt{ab}+2\sqrt{\frac{b}{a}}-\sqrt{\frac{a}{b}}+\frac{1}{\sqrt{ab}}\right).\sqrt{ab}\) (ĐK : \(\hept{\begin{cases}a>0\\b>0\end{cases}}\)hoặc \(\hept{\begin{cases}a< 0\\b< 0\end{cases}}\))
\(=ab+2b-a+1\)
b) \(\left(-\frac{am}{b}\sqrt{\frac{n}{m}}-\frac{ab}{n}.\sqrt{mn}+\frac{a^2}{b^2}.\sqrt{\frac{m}{n}}\right)\left(a^2b^2.\sqrt{\frac{n}{m}}\right)\) (ĐK bạn tự xét nhé ^^)
\(=\left(-\frac{a\sqrt{mn}}{b}-\frac{ab\sqrt{m}}{\sqrt{n}}+\frac{a^2}{b^2}.\sqrt{\frac{m}{n}}\right)\left(a^2b^2.\sqrt{\frac{n}{m}}\right)\)
\(=a^2b^2\left(\frac{-an}{b}-ab+\frac{a^2}{b^2}\right)=-a^3bn-a^3b^3+a^4=a^3\left(a-bn-b^3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
TÍNH : \(\left(\sqrt{2}-1\right)^2-\frac{3}{2}\sqrt{\left(-2\right)^2}+\frac{4\sqrt{2}}{5}+\sqrt{1\frac{11}{25}}.\sqrt{2}\)
\(=\left(\sqrt{2}-1\right)^2-\frac{3}{2}.2+\frac{4\sqrt{2}}{5}+\sqrt{\frac{36}{25}}.\sqrt{2}\)
\(=3-2\sqrt{2}-3+\frac{4\sqrt{2}}{5}+\frac{6\sqrt{2}}{5}=\frac{10\sqrt{2}}{5}-2\sqrt{2}=2\sqrt{2}-2\sqrt{2}=0\)
CHỨNG MINH :
Ta có : \(\sqrt{x}\left(1-\sqrt{x}\right)=-x+\sqrt{x}=-\left[\left(\sqrt{x}\right)^2-2.\sqrt{x}.\frac{1}{2}+\frac{1}{4}\right]+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)với mọi \(x\ge0\)
Vậy ta có điều phải chứng minh.
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
\(\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+\sqrt{7}\right):\sqrt{7}=\left(\frac{1}{\sqrt{7}}-\frac{4}{\sqrt{7}}+\sqrt{7}\right).\frac{1}{\sqrt{7}}=\left(\frac{-3}{\sqrt{7}}+\sqrt{7}\right).\frac{1}{\sqrt{7}}=\frac{-3}{7}+1=\frac{4}{7}\)
Trước hết ta chứng minh bất đẳng thức tổng quát : với n là là số tự nhiên lớn hơn 1 thì :
\(2\sqrt{n-2< 1+1\sqrt{2}+1\sqrt{3}+....+1\sqrt{n}< 2\sqrt{n}-12n-2< 1+12+13+...+1n< 2n-1\left(\cdot\right)\left(\cdot\right)}\)Xét số hạng thứ kk trong dãy : (2 bé hơn hoặc k bé hơn hoặc bằng n ).(2 bé hơn hoặc bằng k bé hơn hoặc bằng n )
Ta có : \(1\sqrt{k>2\sqrt{k}+\sqrt{k}+1=2\left(\sqrt{k}+1-\sqrt{k}\right)1k>2k+k+1=2\left(k+1-k\right)v\text{à}}1\sqrt{k}< 2\sqrt{k}+\sqrt{k}-1=2\left(\sqrt{k}-\sqrt{k}-1\right)1k< 2k+k-1\)\(=2\left(k-k-1\right)\)
Do đó : \(1+1\sqrt{2}+...+1\sqrt{n}>2\left(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+....+\sqrt{n}+1-\sqrt{n}\right)=2\left(\sqrt{n}+1-1\right)>2\sqrt{n}-21+12+.....+1n\)\(>2\left(2-1+3-2+...+n+1-n\right)=2\left(n+1-1\right)>2n-2v\text{à}1+1\sqrt{2}+.....+1\sqrt{n}< 1+2\left(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{n}-\sqrt{n}-1\right)\)\(=1+2\left(\sqrt{n}-1\right)=2\sqrt{n}-11+12+...+1n< 1+2\left(2-1+3-2+...+n-n-1\right)=1+2\left(n-1\right)=2n-1\)Đến đây áp dụng (*)(*) với n=100n=100 thì 19<a<2019<a<20 nên a không phải là số tự nhiên
bào này mình làm hơi mệt đó , sao nó dài quá